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28 votes

In what sense is the proper/effective action $\Gamma[\phi_c]$ a quantum-corrected classical action $S[\phi]$?

There is already a good answer by Solenodon Paradoxus. Here we provide a formal proof (via the stationary phase/WKB approximation). To fix notation, we define the 1PI effective/proper action $$ \...
Qmechanic's user avatar
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25 votes
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In what sense is the proper/effective action $\Gamma[\phi_c]$ a quantum-corrected classical action $S[\phi]$?

We want to calculate the path integral $$ Z = \int \mathcal{D}{\phi}\, e^{i \hbar^{-1} S[\phi]} $$ which encodes a transition amplitude between initial and final quantum states. If we had the ...
Prof. Legolasov's user avatar
23 votes
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Does universal speed limit of information contradict the ability of a particle to pick a trajectory using Principle of Least Action?

The key to this is that the Lagrangian cannot be just any old function. It has to be a function such that, when the action is stationary, its solution describes the kinematics of the system. Thus, if ...
Cort Ammon's user avatar
21 votes

In what sense (if any) is Action a physical observable?

The book propagates a myth. Experiments measure angular momentum, not action - even though these have the same units. One finds empirically that angular momentum in any particular (unit length) ...
Arnold Neumaier's user avatar
21 votes

Are all Lagrangians translationally invariant?

When you write $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x))$, you're assuming translational invariance. A more general Lagrangian is written $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\...
Jahan Claes's user avatar
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20 votes
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In a Lagrangian, why can't we replace kinetic energy by total energy minus potential energy?

OP is essentially asking: Why can't we replace the Lagrangian $L=T-V$ with $L=E-2V$ by using energy conservation $T+V=E$, where $E$ is an integration constant? Answer: Generically an action ...
Qmechanic's user avatar
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18 votes
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The dynamical variables in Lagrangian formalism

We do not treat $\dot q$ as an independent variable in the derivation of the Euler-Lagrange equations. The rough answer is that $q$ and $\dot q$ are independent as inputs to the Lagrangian, but ...
J. Murray's user avatar
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17 votes
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Does classical electrodynamics have a Lagrangian that gives both the Lorentz force and Maxwell equations?

Is there a combined Lagrangian that gives both the Lorentz force and Maxwell equations via the Euler-Lagrange equations? There is the standard textbook Lagrangian for $N$ charged particles: $$ L(A^\...
Ján Lalinský's user avatar
16 votes
Accepted

Has anyone ever come up with a formulation of quantum mechanics in which the action is an operator?

What would come the most close to that would be Julian Schwinger's Quantum Action Principle, see for example this pdf: https://arxiv.org/abs/1503.08091. Or see his paper "the Theory of quantized ...
Quantumwhisp's user avatar
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15 votes

Derivation of Maxwell's equations from field tensor lagrangian

Although late in the party, I post an answer on an elemementary level. May be this proves the power of tensor calculus used in all previous nice answers. Abstract In this answer we'll try to ...
Frobenius's user avatar
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15 votes
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Mass Dimension of derivative in a Lagrangian

We work in units wherein $c = \hbar = 1$. Recall the Compton wavelength is, $$\lambda = \frac{\hbar}{mc}$$ and thus in our units length scales $\lambda$ have units of inverse mass or equivalently ...
JamalS's user avatar
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13 votes

Why the Principle of Least Action?

Here is, at long last, my own attempt to thwack at this problem. To understand it, we first need to, as with many things, take a bit of a step back. We should not so much at first be interested in ...
The_Sympathizer's user avatar
13 votes

Why does the 'metric Lagrangian' approach appear to fail in Newtonian mechanics?

The problem with your approach is that your proposed action $$S = \int |\mathbf{v}| \, dt$$ is not invariant at all. While Landau's action is invariant under Lorentz transformations, and in fact ...
knzhou's user avatar
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13 votes

Does universal speed limit of information contradict the ability of a particle to pick a trajectory using Principle of Least Action?

About global statement and local statement: There is a lemma in variational calculus that was first stated by Jacob Bernoulli. When Johann Bernoulli had presented the Brachistochrone problem to the ...
Cleonis's user avatar
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12 votes

Derivation of Euler-Lagrange equations for Lagrangian with dependence on second order derivatives

Suppose now the Lagrangian $L$ is a function of $y(x), y'(x)$ and also $y''(x)$, i.e. it contains second derivatives w/r to the parameter $x$. It is straightforward to adapt the usual procedure to ...
ZeroTheHero's user avatar
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12 votes
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Is the Lagrangian approach essentially a 'theory of everything'?

The Lagrangian and Hamiltonian approaches are frameworks, and not theories. It is certainly true that a wide variety of systems are susceptible to such an approach. However, there are many theories ...
JamalS's user avatar
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12 votes

Transition from quantum to classical mechanics

The heuristic that compares the action $S$ to Planck's constant is vaguely useful as an initial criterion, but the limit from quantum to classical mechanics is rather more subtle, in ways that make ...
Emilio Pisanty's user avatar
12 votes

Invariance of Action vs. Lagrangian in Noether's theorem?

No, they are not the same. To see why, even in classical mechanics, suppose we have symmetry transformation $q \rightarrow q + \epsilon K$ that leaves the Lagrangian invariant. This means that we must ...
Giuseppe Rossi's user avatar
12 votes
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Can we have conservation of momentum without conservation of energy?

Just adding a function $V(t)$ to the Hamiltonian does nothing - the equations of motion involve only the derivatives of the Hamiltonian w.r.t. $q$ and $p$, and so this changes nothing about the system,...
ACuriousMind's user avatar
  • 124k
11 votes

In the Principle of Least Action, how does a particle know where it will be in the future?

The particle doesn't have to "know" anything. The principle of least action is used when we already know the endpoints of the path, and we want to find out how the particle got from the initial to the ...
knzhou's user avatar
  • 102k
11 votes

Why the Lagrangian $L$ is KE - PE? Why not KE + PE!

There is no intuitive way to grasp this because it is not true! The Lagrangian is NOT the energy of the system. The energy of the system is $(KE+PE)$, of course. I can define lots of quantities with ...
Jahan Claes's user avatar
  • 8,005
11 votes

Why are the Nambu-Goto action and Polyakov action equivalent at quantum level?

The path integral involving the Nambu-Goto square root in the exponent is a very complex animal. Especially in the Minkowski signature, there is no totally universal method to define or calculate the ...
Luboš Motl's user avatar
11 votes

Why are the Nambu-Goto action and Polyakov action equivalent at quantum level?

I) Recall that the path integral formulation comes in (at least) two versions: Lagrangian & Hamiltonian. It is often argued that the Hamiltonian version is more fundamental, cf. e.g. this Phys.SE ...
Qmechanic's user avatar
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11 votes

Is Action Always "Locally" Least?

Introduction. It is well-known that the stationary solution to a simple harmonic oscillator (SHO) (with Dirichlet boundary conditions) is not a local minimum but a saddle point beyond the first ...
Qmechanic's user avatar
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10 votes

Why is action a functional of $q$ only?

The notation $q$ in the functional $S[q]$ stands for the whole parametrized curve/path $q:[t_i,t_f]\to \mathbb{R}$, not just a single position. In particular, the parametrized path already carries all ...
Qmechanic's user avatar
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10 votes

Does Newtonian $F=ma$ imply the least action principle in mechanics?

Newton's second law implies into Least Action Principle under two assumptions: Virtual Work Principle holds. There are no dissipative forces. When applied to a system of particles, Newton´s second ...
Diracology's user avatar
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10 votes
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Is Hilbert-Einstein action just the leading order of some kind of series?

Whenever you talk about "validity" of a theory/equation in physics, you must also specify the length scales being probed. This is important because physics at different scales is different, with ...
Avantgarde's user avatar
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9 votes

How do I show that there exists variational/action principle for a given classical system?

Remark: I have decided to rework this answer completely. I was unhappy with it and it was incomplete. I have replaced functional derivatives as the main tool with Euler and Helmholtz operators as it ...
Bence Racskó's user avatar
9 votes
Accepted

Why is the action dimensionless in natural units?

It is dimensionless in the sense of mass dimension. Setting $\hbar = c = 1$ means we only need to fix one base unit, which is usually taken to be the energy measured in $\mathrm{eV}$. Now, since $c=1$...
ACuriousMind's user avatar
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